【NTT】loj#6261. 一个人的高三楼

去年看过t老师写这题博客；以为是道神仙题

题目大意

求一个数列的$k$次前缀和。$n\le 10^5$.

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题目分析

【计数】cf223C. Partial Sums 加强版。注意到最后的式子是$f_i=\sum\limits_{j+k=i}pre_j a_k$的样子，因此在预处理$pre_j$之后就是卷积的板子了。

 #include<bits/stdc++.h>
 #define MO 998244353
 const int maxn = ;
 const int inv3 = ;

 int n,len,dt;
 int cov[maxn],a[maxn],f[maxn],pre[maxn],fac[maxn],inv[maxn];
 long long k;

 int qmi(int a, int b)
 {
     int ret = ;
     for (; b; b>>=, a=1ll*a*a%MO)
         if (b&) ret = 1ll*ret*a%MO;
     return ret;
 }
 void NTT(int *a, int opt)
 {
     for (int i=; i<len; i++)
         if (i < cov[i]) std::swap(a[i], a[cov[i]]);
     for (int i=; i<len; i<<=)
     {
         int Wn = qmi(, (MO-)/(i<<));
         if (opt==-) Wn = qmi(inv3, (MO-)/(i<<));
         for (int j=, p=i<<; j<len; j+=p)
         {
             int w = ;
             for (int k=; k<i; k++, w=1ll*w*Wn%MO)
             {
                 int valx = a[j+k], valy = 1ll*w*a[i+j+k]%MO;
                 a[j+k] = (valx+valy)%MO, a[i+j+k] = (valx-valy+MO)%MO;
             }
         }
     }
     if (opt==-){
         int inv = qmi(len, MO-);
         for (int i=; i<len; i++) a[i] = 1ll*a[i]*inv%MO;
     }
 }
 void init()
 {
     pre[] = fac[] = fac[] = inv[] = inv[] = ;
     for (int i=; i<=n; i++)
         fac[i] = 1ll*fac[i-]*i%MO,
         inv[i] = MO-1ll*(MO/i)*inv[MO%i]%MO;
     for (int i=; i<=n; i++)
         pre[i] = 1ll*pre[i-]*(k%MO+i-)%MO*inv[i]%MO;
 }
 int main()
 {
     freopen("loj6261.in","r",stdin);
     freopen("loj6261.out","w",stdout);
     scanf("%d%lld",&n,&k);
     for (int i=; i<n; i++) scanf("%d",&a[i]);
     if (k){
         init();
         for (len=; len<=n+n; len<<=) ++dt;
         for (int i=; i<len; i++)
             cov[i] = (cov[i>>]>>)|((i&)<<(dt-));
         NTT(a, ), NTT(pre, );
         for (int i=; i<len; i++) f[i] = 1ll*a[i]*pre[i]%MO;
         NTT(f, -);
     }else for (int i=; i<n; i++) f[i] = a[i];
     for (int i=; i<n; i++) printf("%d\n",f[i]);
     return ;
 }

END